Question

When 12x² + kx + 3 and 3x³ + (3k + 2)x² + 28x +17

are divided by 3x + 2, their remainders are
equal. Find the value of the constant k.​

Answer 1

`\boxed{\sf \ \ \ k=5 \ \ \ }`

Explanation:

Hello,

We know that for a real, the polynomial P(x) divided by (x-a) the remainder is P(a)

and as

`3x+2=0<=>3x=-2<=>x=(-2)/(3)`

so we need

`12((-2)/(3))^2+k((-2)/(3))+3=3((-2)/(3))^3+(3k+2)((-2)/(3))^2+28(-2)/(3)+17\\ \ multiply \ by \ 9\\<=> 12*4-6k+27=-8+4(3k+2)-28*2*3+17*9\\<=> -6k+48+27=-8+12k+8-168+153\\<=> -6k+75=12k-15\\<=> 18k=75+15=90\\<=> k = (90)/(18)=(15)/(3)`

hope this helps